Calculating Velocities in One-Dimensional Collisions

In summary, we just discussed a question regarding Conservation of Momentum and a one-dimensional collision. Using the Momentum equation, we were able to calculate the velocity of the 4.00-kg object after the collision to be 3.5m/s. We also calculated the velocity of the center of mass of the two objects before and after the collision to be 2m/s, indicating that momentum is conserved and the velocity of the center of mass remains unaffected in a collision.
  • #1
JamesRV
4
0
We just started learning Conservation of Momentum, and I feel completely lost. This is the following question I was asked:

Homework Statement


A 2.00-kg object moving at 6.00 m/s collides with a 4.00-kg object that is initially at rest. After the collision, the 2.00-kg object moves backward at 1.00 m/s. This is a one-dimensional collision, and no external forces are acting.

a) Calculate the velocity of the 4.00-kg object after the collision.

My attempt was to simply use the Momentum equation:

Homework Equations

& The attempt at a solution[/B]

mA1vA1 + mB1vB1 = mA2vA2 + mB2vB2

So:
vB2 = (mA1vA1 + mB1vB1 - mA2vA2)/mB2

Resulting in the velocity to be 3.5m/s.

I was just looking for someone to confirm that answer, because I am unsure if I did it right.

In question b) it asks:
Calculate the velocity of the centre of mass of the two objects (i) before the collision and (ii) after the collision.
What does that mean?

Thank you for helping.
 
Last edited:
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  • #2
JamesRV said:
We just started learning Conservation of Momentum, and I feel completely lost. This is the following question I was asked:

A 2.00-kg object moving at 6.00 m/s collides with a 4.00-kg object that is initially at rest. After the collision, the 2.00-kg object moves backward at 1.00 m/s. This is a one-dimensional collision, and no external forces are acting.

a) Calculate the velocity of the 4.00-kg object after the collision.

My attempt was to simply use the Momentum equation:

mA1vA1 + mB1vB1 = mA2vA2 + mB2vB2

So:
vB2 = (mA1vA1 + mB1vB1 - mA2vA2)/mB2

Resulting in the velocity to be 3.5m/s.

I was just looking for someone to confirm that answer, because I am unsure if I did it right.

In question b) it asks:
Calculate the velocity of the centre of mass of the two objects (i) before the collision and (ii) after the collision.
What does that mean?

Thank you for helping.

Hi JamesRV

Welcome to PF!

The result of part a) looks right .

For part b) use Vcm = (m1v1+m2v2)/(m1+m2) .

Apply this before and after collision .What do you get ?

Please take care of the signs .
 
Last edited:
  • #3
Thank you Tanya!

I used the equation you gave me, it makes logic sense taking the total momentum and dividing by the mass.
I got 2m/s for both cases, which seems right!
 
  • #4
JamesRV said:
Thank you Tanya!

I used the equation you gave me, it makes logic sense taking the total momentum and dividing by the mass.
I got 2m/s for both cases, which seems right!

Yes..Well done :thumbs:

So what does this tell you about collision ?
 
  • #5
Tanya Sharma said:
Yes..Well done :thumbs:

So what does this tell you about collision ?

Doesn't it just mean that the momentum is conserved? I thought momentum was always conserved, is it not?
 
  • #6
JamesRV said:
Doesn't it just mean that the momentum is conserved? I thought momentum was always conserved, is it not?

Yes... And the velocity of Center of Mass remains unaffected in a collision.
 
  • #7
Okay, thanks again!
 

FAQ: Calculating Velocities in One-Dimensional Collisions

1. What is conservation of momentum?

Conservation of momentum is a fundamental principle in physics that states that the total momentum of a closed system remains constant in the absence of external forces.

2. How is conservation of momentum related to Newton's laws of motion?

Newton's third law of motion, which states that for every action there is an equal and opposite reaction, is directly related to conservation of momentum. This is because when two objects interact, the total momentum of the system is conserved.

3. What are the types of collisions in which conservation of momentum applies?

Conservation of momentum applies to both elastic and inelastic collisions. In elastic collisions, both the total momentum and kinetic energy of the system are conserved. In inelastic collisions, only the total momentum is conserved.

4. How can conservation of momentum be applied to real-world scenarios?

Conservation of momentum can be applied to various real-world scenarios, such as in car crashes, where the total momentum of the system before and after the collision remains constant. It is also used in rocket propulsion, where the momentum of the exhaust gas propels the rocket forward.

5. What is the formula for calculating momentum in conservation of momentum problems?

The formula for calculating momentum is mass multiplied by velocity (p = mv). In conservation of momentum problems, the total momentum of the system before and after the interaction must be equal.

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